Thomas Gummerer <[email protected]> writes:
> Since the cache-tree data is saved as part of the directory data,
> we have already read it, when we want to read the cache-tree. The
> cache-tree then only has to be converted from the directory data.
I think the first sentence is wrong. You have already read it at the
very beginning of reading the index format, when you parsed the
directory records, haven't you?
> The cache-tree isn't lexically sorted, but after the pathlen at
> each level, therefore the directories have to be reordered with
> respect to the ondisk layout.
I'm not a native speaker either, but I think this does't parse well.
Maybe
The cache-tree data is arranged in a tree, with the children sorted by
pathlen at each node. So we have to rebuild this format from the
on-disk directory list.
> + for (i = 0; i < subtree_nr; i++) {
> + struct cache_tree *sub;
> + struct cache_tree_sub *subtree;
> + char *buf, *name;
> +
> + name = "";
> + buf = strtok(down[i].de->pathname, "/");
man 3 strtok says
Be cautious when using these functions. If you do use them, note
that:
* These functions modify their first argument.
* These functions cannot be used on constant strings.
* The identity of the delimiting character is lost.
* The strtok() function uses a static buffer while parsing, so it's
not thread safe. Use strtok_r() if this matters to you.
I don't think the last point will be a problem, but what about modifying
the argument?
--
Thomas Rast
trast@{inf,student}.ethz.ch
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